overloading wirewound resistors

Insignificant for DC driven magnetic coils.

Why do you make this stuff up, when there all that stuff is available on Google? Gradient coils are *not* DC driven.

The purpose of the troublesome boost supply is to bang the L di/dt of the coil inductance on sharp rising edges. Peak voltage requirements for a flat pulse are typically twice the steady-state value. L's run in the 10's of microhenries, sometimes more.

John

firmware

handle

existing

hurricane

pushed

experience?

ask.

connection

lugs

all

In the end, the Proctologist has the final say.

I am very reluctant to stick my less than $.02 in, since this was a problem I dealt with in '78. Never much call in worrying about this when I was architecting computers after that time. What triggered my thoughts is your way of describing the resistors as exploding. Not overheating. The issue as it was explained to me by the resistor rep when we were building lightning simulators is the joule rating of the resistors, not the wattage. Some resistors are specifically designed to handle short bursts of energy, see the description here:

If this is the problem, then you can chill it anyway that you want, and it will not solve the problem of the resistors exploding. We had to use a large number of high energy resistors in parallel to keep from spraying resistor shards across the lab.

So I am probably off base, but it might be worth considering if they are really exploding without overheating.

-Jim

It was all explained in the Harris white papers, but the basics are: A combiner in each module tray that does impedance matching, and more of the same are in the racks that hold the module trays. It is similar to how a CATV headend combines dozens or a hundred channels, yet maintains the required 75 ohm impedance all through the system. the ports are backmatched, and unused ports are terminated.

I may have copies of most of the harris Broadcast white papers on an old computer, but it isn't set up right now.

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Yes, it is variable DC, but the lost due to induction is insignificant as compared to resistive lost. You can say it is AC.

can't say it's AC.

microHenries are not big induction. Yes, the problem is you have to change current very fast. But the main energy lost (and heat generation) is still in the resistive element.

So it looks like an inductor above 1KHz?

RL

My

stuff,

I said it's a complex waveform. Ever been MRI'd? It's very noisy.

Sure. It always is.

John

human

adequate. My

stuff,

Not much energy is induced into the target. Otherwise, I wouldn't go near that giant cooker. In fact, they might be trying to reduce induction by just varying the fields rather than pulsing. That might explains why you are seeing such high duty cycles.

....if the source is loss-free. I understood that this source is linearly regulated, and that there was a problem with internal elements overheating........

RL

I was thinking along the same lines as Ecnerwal, but where he suggested machining some steel, I was thinking rather more amateurishly:

20 ohms is probably feasible to create as PCB tracks. Use 2 oz copper on a multilayer PCB, so the internal planes can help heatsink it. Put vanes on either side of the track, where you can attach heatsinks. (Or distribute the resistance with lots of fine parallel tracks, giving them a larger area of substrate to dissipate across.) Run the track along one side, then mirror the return path along the other side, so it's essentially non-inductive.

(I've always assumed you can get solder-on heatsinks for surface mounting on a PCB, but never actually seen any. They'd be efficient at getting the heat off for this solution.)

Whether the substrate (FR4?) and the glue bonding the copper to the board can take the thermal stresses is a moot point. As is whether the tracks would just burn out. But you say you have a massive airflow and PCB's can be very multipurposed, i.e. it's easy to add convenient mounting holes etc.

The advantage of this approach is that the heat would be dissipated over a pretty large area of copper, so the thermal gradients might not be too massive, which seems to be what's disintegrating your resistors. Joule rating, as someone said.

--
Nemo

to the OP..

maybe you can come up with an R C circuit, maybe throw a diode or two in for good measure, that can smooth out and reduce the PEAK current through the resistors? I don't think you can change the energy lost per pulse (unless you change the input and/or output voltage) but you may be able to smooth out the current and reduce the peak power dissipation which seems to be the issue....

Mark

Interim results:

A 25 watt vitreous ww resistor can handle maybe 3x rated power at 700 LFPM airflow, hitting maybe 350C surface temp. At about 9x, maybe

850C, it melts:

ftp://jjlarkin.lmi.net/WW_RES_MELTED.JPG

although it's still the original resistance.

Now these are slick:

ftp://jjlarkin.lmi.net/Welwyn.JPG

Porcelain and thickfilm over spring steel. It's slightly curved, so the single screw pushes it flat against most any surface. This one is

2/3 the size of a business card and is rated at 780 watts on a heatsink, although I'd be afraid of melting the faston solder at that power density. They're cheap, around $7, a real deal for that kind of power.

John

Be careful with molten glass. Burned myself _once_ like that... most painful burn I've ever endured.

Cool! While hot ;-)

...Jim Thompson

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Put two with twice the requisite R value in parallel. Or half the value in series.

With the proper heat sink, the above parts are good for 600W each.

--
Paul Hovnanian     mailto:Paul@Hovnanian.com
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together all the time. We called it algebra class. -- Jay Leno

If what I read in a previous reply is true (the waveform having a low duty cycle), that isn't going to help much. The heatsinking, airflow, etc. is only going to carry off the average power.

The only thing that will help you with high peak power dissipation is to increase the mass of the current carrying parts. Or their melting temperature.

It shouldn't be too difficult to do a spice model of the problem, converting the thermal coefficients into equivalent electrical ones and solving for the peak values.

--
Paul Hovnanian  paul@hovnanian.com
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Have gnu, will travel.